sin6a +cos6a +3sin2a.cos2a= 1
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Answered by
55
L.H.S = sin6a+cos6a+3sin2acos2a= sin6a+cos6a+3sin2acos2a(sin2a+cos2a) {Since, (sin2a+cos2a) = 1}= (sin2a)3+(cos2a)3+3sin2acos2a(sin2a+cos2a)= (sin2a+cos2a)3 {Since; a3+b3+3ab(a+b) = (a+b)3}= 13= 1= R.H.SHence proved.
Answered by
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yanna I will help you
sin6A + cos6A = (sin^2A )^3+ (cos^2A)^3
= (sin^2A + cos^2A) (sin^4A - sin^2A cos^2A +cos^4A) (using identity (a3+ b3= (a+b) (a2 + ab + b2 )
= (1) (sin^2A)^2 +2 sin^2A cos^2A - 2sin^2 A cos^2A -sin^2A cos^2A + cos^4A)
= (sin^2A + cos^2A)^2- 3 sin^2A cos^2A
= 1- 3 sinA^2 cosA^2
hmmm
sin6A + cos6A = (sin^2A )^3+ (cos^2A)^3
= (sin^2A + cos^2A) (sin^4A - sin^2A cos^2A +cos^4A) (using identity (a3+ b3= (a+b) (a2 + ab + b2 )
= (1) (sin^2A)^2 +2 sin^2A cos^2A - 2sin^2 A cos^2A -sin^2A cos^2A + cos^4A)
= (sin^2A + cos^2A)^2- 3 sin^2A cos^2A
= 1- 3 sinA^2 cosA^2
hmmm
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